# Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 3

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## Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let each point $x$ of $T$ have a local basis $\mathcal D_x$ consisting entirely of path-connected sets in $T$.

Then

- $T$ has a basis consisting of path-connected sets in $T$.

## Proof

By definition of local basis, each of these path-connected sets in $\mathcal D_x$ is open in $T$.

Consider the set $\displaystyle \mathcal D = \bigcup_{x \mathop \in S} \mathcal D_x$.

From Union of Local Bases is Basis, $\mathcal D$ is a basis for the topology $\tau$.

Since each $\mathcal D_x$ consists entirely of path-connected sets, $\mathcal D$ also consists entirely of path-connected sets.

$\blacksquare$